Sinopsis
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1905 Excerpt: ...in two sets. In one, call it V0, we put the points for which Ax = 0. Then Vo = ± ± 1 ni m + 1 In the other set, call it Vu we put all the other points of V. We can now show for the function y = f(x) in the above theorem, that 1) is true for each one of these sets of points, and therefore true for both together. 379. We give now the proof of 378. Let t be any point in T let x he the corresponding point in X. Let As, Ay he the increments of x, y, corresponding to the increment At of t. Case 1. There exists a V(t), in which AxO. The identity Ay = Ay At,j At Ax ' At does not involve a division by 0, as Ax 0. which proves the theorem for this case. Case 2. Ax = 0 for some point in every V(t. Let FJ, be the points of V(t), for which Ax = 0. Let Fj be the remaining points of V(t). If we show linif, and lim, (3 At dx At have one and the same value for every sequence of points whose limit is t, we have proved 2) for this case. Let A be any sequence in V0. Then 11=0, (4 At For, Ax being 0 for every point of A, y=f(x) receives no increment, and hence Ay = 0 in A. Thus, for every sequence A, the two limits in 3) have the same value, viz. 0. Let now B be any sequence in Vl which = t. Let the image of the points B be the points C, on the z-axis. Then, by 292, lim--f = lim lim-r-r j At o Ax « At ax B At Thus the two limits of 3) are the same for each sequence B. It remains to show that one is 0. Now, by 4), for any sequence whose limit is the point f. Hence the right side of 5) is 0. Thus the two limits 3) have the value 0 for every sequence A or B. These limits therefore have the value 0 for any sequence, whether its points all lie in V0, or in Vv or partly in V0 and partly in Vy 380. The demonstration, as ordinarily given, rests on the identity... Ay _ Ay...
Reseña del editor
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1905 Excerpt: ...in two sets. In one, call it V0, we put the points for which Ax = 0. Then Vo = ± ± 1 ni m + 1 In the other set, call it Vu we put all the other points of V. We can now show for the function y = f(x) in the above theorem, that 1) is true for each one of these sets of points, and therefore true for both together. 379. We give now the proof of 378. Let t be any point in T let x he the corresponding point in X. Let As, Ay he the increments of x, y, corresponding to the increment At of t. Case 1. There exists a V(t), in which AxO. The identity Ay = Ay At,j At Ax ' At does not involve a division by 0, as Ax 0. which proves the theorem for this case. Case 2. Ax = 0 for some point in every V(t. Let FJ, be the points of V(t), for which Ax = 0. Let Fj be the remaining points of V(t). If we show linif, and lim, (3 At dx At have one and the same value for every sequence of points whose limit is t, we have proved 2) for this case. Let A be any sequence in V0. Then 11=0, (4 At For, Ax being 0 for every point of A, y=f(x) receives no increment, and hence Ay = 0 in A. Thus, for every sequence A, the two limits in 3) have the same value, viz. 0. Let now B be any sequence in Vl which = t. Let the image of the points B be the points C, on the z-axis. Then, by 292, lim--f = lim lim-r-r j At o Ax « At ax B At Thus the two limits of 3) are the same for each sequence B. It remains to show that one is 0. Now, by 4), for any sequence whose limit is the point f. Hence the right side of 5) is 0. Thus the two limits 3) have the value 0 for every sequence A or B. These limits therefore have the value 0 for any sequence, whether its points all lie in V0, or in Vv or partly in V0 and partly in Vy 380. The demonstration, as ordinarily given, rests on the identity... Ay _ Ay...
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